**11. Evaluate **

**The sum of first three terms of a G.P. is**^{39}/_{10}and their product is 1. Find the common ratio and the terms.

Solution:

Let ^{a}/_{r}, a, ar be the first three terms of the G.P.

^{a}/_{r} + a + ar = ^{39}/_{10} —-(1)

(^{a}/_{r})(a)(ar) = 1 —–(2)

From (2), we obtain a^{3} = 1

a = 1

Substuting a = 1 in equation (1), we get

^{1}/_{r} + 1 + r = ^{39}/_{10}

1 + r + r^{2} = ^{39}/_{10} r

10 + 10r + 10r^{2} – 39r = 0

10r^{2} – 29r + 10 = 0

10r^{2} – 25r – 4r + 10 = 0

5r(2r – 5)-2(2r – 5)= 0

(5r – 2)(2r – 5) = 0

r = ^{2}/_{5} or ^{5}/_{2}

Thus the three terms of G.P. are ^{5}/_{2}, 1 and ^{2}/_{5}

**How many terms of G.P. 3, 3**^{2},3^{3}… are needed to give the sum 120?

Solution:

The given G.P. is 3, 3^{2}, 3^{3},…

Let n terms of this G.P. be required to obtain the sum as 120.

Thus, four terms of the given G.P. are required to obtain the sum as 120.

**14: The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. ****Determine the first term, the common ratio and the sum of n terms of the G.P.**

Solution:

Let the G.P. be a, ar, ar^{2},…

According to the given condition,

a + ar + ar^{2} = 16 and ar^{3} + ar^{4} + ar^{5} = 128

⇒ a (1 + r + r^{2}) = 16 ……………. (1)

ar^{3}(1 + r + r^{2}) = 128 …………….. (2)

Dividing equation (2) by (1), we obtain

**Given a G.P. with a = 729 and 7**^{th}term 64, determine S_{7}

Solution:

a = 729 and a_{7= }64

Let r be the common ratio of the G.P. It is known that

a_{n} = a.r^{n-1}

a_{7} = a.r^{7-1} =(729)r^{6}

64 = 729(r^{6})

r^{6} = ^{64}/_{729}

r^{6} =(^{2}/_{3})^{6}

r = ^{2}/_{3}

Also it is known that,

**Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.**

Solution:

Let a be the first term and r be the common ratio of the G.P.

According to the given conditions,

Thus the required G.P. is –^{4}/_{3}, –^{8}/_{3}, –^{16}/_{3},… or 4, -8, 16, -32,…

**If the 4**^{th},10^{th}and 16^{th}terms of a G.P. are x , y and z respectively. Prove that x , y, z are in G.P.

Solution:

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a_{4} = ar^{3} = x ———-(1)

a_{10} = ar^{9} = y ———-(2)

a_{16} = ar^{15} = z ———-(3)

Dividing (2) by (1), we get

^{y}/_{x} = (ar^{9})/(ar^{3})

^{y}/_{x} = r^{6}

Dividing (3) by (2), we get

^{z}/_{y} = (ar^{15})/(ar^{9})

^{z}/_{y} = r^{6}

^{y}/_{x} =^{ z}/_{y}

Thus x, y, z are in G.P.

**Find the sum to n terms of the sequence, 8, 88, 888, 8888…**

Solution:

The given sequence is 8, 88, 888, 8888…

This sequence is not a G.P. However, it can be changed to G.P. as

Sn = 8 + 88 + 888 + 8888 + …………….. to n terms

= ^{8}/_{9} [9 +99 + 999+ 9999+ … to n terms]

= ^{8}/_{9} [(10-1) +(100-1) + (1000-1) + (1000-1) + … to n terms]

= ^{8}/_{9} [(10 +10^{2} + 10^{3} + … to n terms)-(1+1+1+…n terms)]

= ^{8}/_{9} [^{10(10^n-1)}/_{9} – n]

= ^{80}/_{81} (10^{n} – 1) – ^{8}/_{9 }n

**Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,**^{1}⁄_{2}.

Solution:

Required sum = 2×128 + 4×32 + 8×8 +16×2 +32x^{1}⁄_{2}

=64[4+2+1+^{1}/_{2} +^{1}/_{2}^{2}]

Here 4, 2, 1, ^{1}/_{2}, ^{1}/_{2}^{2} is a G.P.

First term a = 4

Common ratio r = ^{1}/_{2}

Required sum = 64(^{31}/_{4}) = 16(31) = 496

**Show that the products of the corresponding terms of the sequnnces form a, ar, ar**^{2},…ar^{n-1}and A, AR,AR^{2},…,AR^{n-1}a G.P. and find the common ratio.

Solution:

It has to be proved that the sequence: aA, arAR, ar^{2}AR^{2},…ar^{n-1}AR^{n-1} forms a G.P.

^{Second term}/_{first term} = ^{arAR}/_{aA} = rR

^{Third term }/_{secondterm} = ar^{2}AR^{2}/_{arAR} = rR

Thus the above sequence forms a G.P. and the common ratio is rR

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